How evaluate $ \int_0^\infty \frac{\ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|}{1+x^2} \;dx$ How evaluate (without using Complex analysis)
$$ \int_0^\infty \frac{\ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|}{1+x^2}\; dx\quad (a\gt0)$$
My Attempt:
I used the expansion of the following functions: $$ \ln\left|2\sin(x)\right| \text{ and }\ln\left|2\cos(x)\right| $$
To get the following expansion: $$ \ln\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|=2\sum_{n=1}^\infty (-1)^n \frac{\sin[(2n-1)ax]}{2n-1} $$
Then I expressed the square of the logarithmic function as follows:$$ \ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|=4\sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{m+n} \frac{\sin[(2m-1)ax]\sin[(2n-1)ax]}{(2m-1)(2n-1)} $$
And used the formula of the product of two sines, then integrated the following well known integral under the summation sign:
$$ \int_0^\infty \frac{\cos(bx)}{1+x^2} \;dx$$
And expressed the final result in terms of $\; \arctan\;\left(\;e^{-a}\;\right)\; $
but this was not the right answer according to the one evaluated by Complex analysis, which is
$$ \frac{{\pi}^3}{8}-2\pi\; \arctan^2\;\left(\;e^{-a}\;\right) $$
Any hint for another method or idea?
 A: Using OP's series expansion, we have
\begin{align*}
I
&= 2 \sum_{m, n \geq 0} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} \int_{0}^{\infty} \frac{\sin[(2m+1)ax]\sin[(2n+1)ax]}{1+x^2} \, dx \\
&= \pi \sum_{m, n \geq 0} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} \left( e^{-a|2m-2n|} - e^{-a(2m+2n+2)} \right).
\end{align*}
We will split the sum into several parts and analyze them separately.


*

*It is straightforward that
$$ \sum_{m, n \geq 0} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} e^{-a(2m+2n+2)} = \arctan^2(e^{-a}). $$

*Using the identity $\frac{1}{(2m+1)(2n+1)} = \frac{1}{(2m-2n)(2n+1)} + \frac{1}{(2n-2m)(2m+1)}$, we have
\begin{align*}
\sum_{m, n \geq 0} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} e^{-a|2m-2n|}
&= \sum_{m = 0} \frac{1}{(2m+1)^2} + \sum_{\substack{m, n \geq 0 \\ m \neq n }} \frac{(-1)^{m-n}}{(m-n)(2n+1)} e^{-a|2m-2n|} \\
&= \frac{\pi^2}{8} + \sum_{n \geq 0} \frac{1}{2n+1} \sum_{\substack{k \geq -n \\ k \neq 0}} \frac{(-1)^k}{k} e^{-2a|k|}.
\end{align*}

*The last sum can be simplified further: using the fact that $\frac{(-1)^k}{k}e^{-2a|k|}$ is an odd function of $k$,
\begin{align*}
\sum_{n \geq 0} \frac{1}{2n+1} \sum_{\substack{k \geq -n \\ k \neq 0}} \frac{(-1)^k}{k} e^{-2a|k|}
&= \sum_{n \geq 0} \frac{1}{2n+1} \sum_{k = n+1}^{\infty} \frac{(-1)^k}{k} e^{-2a|k|} \\
&= \sum_{k = 1}^{\infty} \left( \sum_{n=0}^{k-1} \frac{1}{2n+1} \right) \frac{(-1)^k}{k} e^{-2a|k|}.
\end{align*}
Symmetrizing the inner sum, we find that
\begin{align*}
\sum_{n=0}^{k-1} \frac{1}{2n+1}
= \frac{1}{2} \sum_{\substack{i, j \geq 0 \\ i+j = k-1}} \left( \frac{1}{2i+1} + \frac{1}{2j+1} \right)
= \sum_{\substack{i, j \geq 0 \\ i+j = k-1}} \frac{k}{(2i+1)(2j+1)}.
\end{align*}
Plugging this back,
\begin{align*}
\sum_{n \geq 0} \frac{1}{2n+1} \sum_{\substack{k \geq -n \\ k \neq 0}} \frac{(-1)^k}{k} e^{-2a|k|}
&= \sum_{i, j \geq 0} \frac{1}{(2i+1)(2j+1)} (-1)^{i+j+1} e^{-a(2i+2j+2)} \\
&= - \arctan^2(e^{-a}).
\end{align*}
Combining altogether, we obtain the desired answer.
